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sesenic [268]
3 years ago
13

A ball is thrown at an angle of to the ground. If the ball lands 90 m away, what was the initial speed of the ball?

Physics
1 answer:
andre [41]3 years ago
8 0

Answer:

The initial speed of the ball is 30 m/s.

Explanation:

It can be assumed that the ball is thrown at an angle of 45 degrees to the ground. The ball lands 90 m away. We need to find the initial speed of the ball. We know that the horizontal distance covered by the projectile is called its range. It is given by :

R=\dfrac{u^2\ sin2\theta}{g}

u is the initial speed of the ball.

v=\sqrt{\dfrac{Rg}{sin2\theta}}

v=\sqrt{\dfrac{90\times 9.8}{sin2(45)}}

v = 29.69 m/s

or

v = 30 m/s

So, the initial speed of the ball is 30 m/s. Hence, this is the required solution.                                      

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When an object is put into a liquid, it experiences a buoyant force that is equal to the weight of the liquid the object displac
Anon25 [30]

Answer:

A) F = 1.8375 N, B) V = 1.25 10⁻⁴ m³, C) V= 1.25 10⁻⁴ m³

Explanation:

This exercise should select some values ​​such as the weight of the block m=0.250 kg and the density of the liquid ρ = 500 kg / m³ (water); with these values ​​answer the question, for this we use the static equilibrium relations.

          Σ F = 0

          F + B - W = 0

the expression for thrust is

          B = ρ g V_liquid

A) the weight of the block is

          W = m g

          W = 0.250 9.8

          W = 2.45 N

Let's look for the maximum push

           V_body = V_liquid

            B = 500 9.8 0.05³

           B = 0.6125 N

we substitute

          F = W - B

          F = 2.45 - 0.6125

          F = 1.8375 N

B) As the body is totally submerged, the volume of the liquid and the body are equal

           V = l³

           V = 0.05³

           V = 1.25 10⁻⁴ m³

C) the volume of water displaced is equal to the volume of the body

           V_liquid = 1.25 10⁻⁴ m³

7 0
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